3.155 \(\int \log (x) \sin ^2(a+b x) \, dx\)

Optimal. Leaf size=66 \[ \frac{\sin (2 a) \text{CosIntegral}(2 b x)}{4 b}+\frac{\cos (2 a) \text{Si}(2 b x)}{4 b}-\frac{\log (x) \sin (a+b x) \cos (a+b x)}{2 b}-\frac{x}{2}+\frac{1}{2} x \log (x) \]

[Out]

-x/2 + (x*Log[x])/2 + (CosIntegral[2*b*x]*Sin[2*a])/(4*b) - (Cos[a + b*x]*Log[x]*Sin[a + b*x])/(2*b) + (Cos[2*
a]*SinIntegral[2*b*x])/(4*b)

________________________________________________________________________________________

Rubi [A]  time = 0.114001, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.546, Rules used = {2635, 8, 2554, 3303, 3299, 3302} \[ \frac{\sin (2 a) \text{CosIntegral}(2 b x)}{4 b}+\frac{\cos (2 a) \text{Si}(2 b x)}{4 b}-\frac{\log (x) \sin (a+b x) \cos (a+b x)}{2 b}-\frac{x}{2}+\frac{1}{2} x \log (x) \]

Antiderivative was successfully verified.

[In]

Int[Log[x]*Sin[a + b*x]^2,x]

[Out]

-x/2 + (x*Log[x])/2 + (CosIntegral[2*b*x]*Sin[2*a])/(4*b) - (Cos[a + b*x]*Log[x]*Sin[a + b*x])/(2*b) + (Cos[2*
a]*SinIntegral[2*b*x])/(4*b)

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

\begin{align*} \int \log (x) \sin ^2(a+b x) \, dx &=\frac{1}{2} x \log (x)-\frac{\cos (a+b x) \log (x) \sin (a+b x)}{2 b}-\int \left (\frac{1}{2}-\frac{\sin (2 a+2 b x)}{4 b x}\right ) \, dx\\ &=-\frac{x}{2}+\frac{1}{2} x \log (x)-\frac{\cos (a+b x) \log (x) \sin (a+b x)}{2 b}+\frac{\int \frac{\sin (2 a+2 b x)}{x} \, dx}{4 b}\\ &=-\frac{x}{2}+\frac{1}{2} x \log (x)-\frac{\cos (a+b x) \log (x) \sin (a+b x)}{2 b}+\frac{\cos (2 a) \int \frac{\sin (2 b x)}{x} \, dx}{4 b}+\frac{\sin (2 a) \int \frac{\cos (2 b x)}{x} \, dx}{4 b}\\ &=-\frac{x}{2}+\frac{1}{2} x \log (x)+\frac{\text{Ci}(2 b x) \sin (2 a)}{4 b}-\frac{\cos (a+b x) \log (x) \sin (a+b x)}{2 b}+\frac{\cos (2 a) \text{Si}(2 b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0766454, size = 50, normalized size = 0.76 \[ \frac{\sin (2 a) \text{CosIntegral}(2 b x)+\cos (2 a) \text{Si}(2 b x)-\log (x) \sin (2 (a+b x))-2 b x+2 b x \log (x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[x]*Sin[a + b*x]^2,x]

[Out]

(-2*b*x + 2*b*x*Log[x] + CosIntegral[2*b*x]*Sin[2*a] - Log[x]*Sin[2*(a + b*x)] + Cos[2*a]*SinIntegral[2*b*x])/
(4*b)

________________________________________________________________________________________

Maple [C]  time = 0.128, size = 132, normalized size = 2. \begin{align*}{\frac{x\ln \left ( x \right ) }{2}}-{\frac{\ln \left ( x \right ) \sin \left ( 2\,bx+2\,a \right ) }{4\,b}}-{\frac{{{\rm e}^{-2\,ia}}\pi \,{\it csgn} \left ( bx \right ) }{8\,b}}+{\frac{{{\rm e}^{-2\,ia}}{\it Si} \left ( 2\,bx \right ) }{4\,b}}-{\frac{{\frac{i}{8}}{{\rm e}^{-2\,ia}}{\it Ei} \left ( 1,-2\,ibx \right ) }{b}}+{\frac{a\ln \left ( ibx \right ) }{2\,b}}-{\frac{x}{2}}-{\frac{a}{2\,b}}-{\frac{a\ln \left ( a+i \left ( ibx+ia \right ) \right ) }{2\,b}}+{\frac{{\frac{i}{8}}{{\rm e}^{2\,ia}}{\it Ei} \left ( 1,-2\,ibx \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(x)*sin(b*x+a)^2,x)

[Out]

1/2*x*ln(x)-1/4*ln(x)/b*sin(2*b*x+2*a)-1/8/b*exp(-2*I*a)*Pi*csgn(b*x)+1/4/b*exp(-2*I*a)*Si(2*b*x)-1/8*I/b*exp(
-2*I*a)*Ei(1,-2*I*b*x)+1/2/b*a*ln(I*b*x)-1/2*x-1/2*a/b-1/2/b*a*ln(a+I*(I*b*x+I*a))+1/8*I/b*exp(2*I*a)*Ei(1,-2*
I*b*x)

________________________________________________________________________________________

Maxima [C]  time = 1.20782, size = 107, normalized size = 1.62 \begin{align*} \frac{{\left (2 \, b x + 2 \, a - \sin \left (2 \, b x + 2 \, a\right )\right )} \log \left (x\right )}{4 \, b} - \frac{4 \, b x +{\left (i \,{\rm Ei}\left (2 i \, b x\right ) - i \,{\rm Ei}\left (-2 i \, b x\right )\right )} \cos \left (2 \, a\right ) + 4 \, a \log \left (x\right ) -{\left ({\rm Ei}\left (2 i \, b x\right ) +{\rm Ei}\left (-2 i \, b x\right )\right )} \sin \left (2 \, a\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/4*(2*b*x + 2*a - sin(2*b*x + 2*a))*log(x)/b - 1/8*(4*b*x + (I*Ei(2*I*b*x) - I*Ei(-2*I*b*x))*cos(2*a) + 4*a*l
og(x) - (Ei(2*I*b*x) + Ei(-2*I*b*x))*sin(2*a))/b

________________________________________________________________________________________

Fricas [A]  time = 2.21566, size = 211, normalized size = 3.2 \begin{align*} \frac{4 \, b x \log \left (x\right ) - 4 \, \cos \left (b x + a\right ) \log \left (x\right ) \sin \left (b x + a\right ) - 4 \, b x +{\left (\operatorname{Ci}\left (2 \, b x\right ) + \operatorname{Ci}\left (-2 \, b x\right )\right )} \sin \left (2 \, a\right ) + 2 \, \cos \left (2 \, a\right ) \operatorname{Si}\left (2 \, b x\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

1/8*(4*b*x*log(x) - 4*cos(b*x + a)*log(x)*sin(b*x + a) - 4*b*x + (cos_integral(2*b*x) + cos_integral(-2*b*x))*
sin(2*a) + 2*cos(2*a)*sin_integral(2*b*x))/b

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \log{\left (x \right )} \sin ^{2}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(x)*sin(b*x+a)**2,x)

[Out]

Integral(log(x)*sin(a + b*x)**2, x)

________________________________________________________________________________________

Giac [C]  time = 1.31466, size = 166, normalized size = 2.52 \begin{align*} \frac{1}{4} \,{\left (2 \, x - \frac{\sin \left (2 \, b x + 2 \, a\right )}{b}\right )} \log \left (x\right ) - \frac{4 \, b x \tan \left (a\right )^{2} + \Im \left ( \operatorname{Ci}\left (2 \, b x\right ) \right ) \tan \left (a\right )^{2} - \Im \left ( \operatorname{Ci}\left (-2 \, b x\right ) \right ) \tan \left (a\right )^{2} + 2 \, \operatorname{Si}\left (2 \, b x\right ) \tan \left (a\right )^{2} + 4 \, b x - 2 \, \Re \left ( \operatorname{Ci}\left (2 \, b x\right ) \right ) \tan \left (a\right ) - 2 \, \Re \left ( \operatorname{Ci}\left (-2 \, b x\right ) \right ) \tan \left (a\right ) - \Im \left ( \operatorname{Ci}\left (2 \, b x\right ) \right ) + \Im \left ( \operatorname{Ci}\left (-2 \, b x\right ) \right ) - 2 \, \operatorname{Si}\left (2 \, b x\right )}{8 \,{\left (b \tan \left (a\right )^{2} + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)*sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/4*(2*x - sin(2*b*x + 2*a)/b)*log(x) - 1/8*(4*b*x*tan(a)^2 + imag_part(cos_integral(2*b*x))*tan(a)^2 - imag_p
art(cos_integral(-2*b*x))*tan(a)^2 + 2*sin_integral(2*b*x)*tan(a)^2 + 4*b*x - 2*real_part(cos_integral(2*b*x))
*tan(a) - 2*real_part(cos_integral(-2*b*x))*tan(a) - imag_part(cos_integral(2*b*x)) + imag_part(cos_integral(-
2*b*x)) - 2*sin_integral(2*b*x))/(b*tan(a)^2 + b)